Date: 01/06/99 at 09:33:07 From: A.H.Noble Subject: Barycentric Calculus Please give me infomation on the workings of Barycentric Calculus. I first came across the subject in David Wells' book, _You Are a Mathematician_, which gives a tantalizing intoduction, and I would like to know more.
Date: 01/08/99 at 08:07:22 From: Doctor Floor Subject: Re: Barycentric Calculus Hello! Thank you for sending your question to Dr. Math. Barycentric coordinates (or "barycentrics") are coordinates with respect to a reference triangle ABC. The original interpretation of the point P given by barycentric coordinates p:q:r is that P is the center of mass ("barycenter") of weights p, q and r when these are placed on the vertices A, B and C. Another interpretation is that the ratio of the areas BCP:ACP:ABP is p:q:r. These interpretations are equivalent. It is also to be understood that for any nonzero real t p:q:r and tp:tq:tr represent the same point. For this reason barycentric coordinates are said to be "homogeneous." Another example of homogeneous coordinates with respect to a triangle is the notion of trilinear coordinates. Trilinear coordinates are given by the ratio p:q:r of the distances to the sidelines of the reference triangle. A short introduction on trilinear coordinates (trilinears) can be found at: http://faculty.evansville.edu/ck6/tcenters/trilin.html Barycentrics have a lot of advantages over trilinear. They have been discussed in the geometry.research newsgroup. You can find the most interesting letter from the discussion, for the largest part consisting of a quote by Professor John Conway, at: http://mathforum.org/kb/message.jspa?messageID=1091956 (Select 30 Jun 1998, "Conway on Trilinear vs Barycentric coordinates, by steve sigur.") When a point has trilinears t:u:v then its barycentrics are at:bu:cv, where a,b,c are the sidelengths of the reference triangle. The barycentric coordinates of the vertices of the reference triangle are simple: A 1:0:0 B 0:1:0 C 0:0:1 Three points given by a1:b1:c1, a2:b2:c2 and a3:b3:c3 are collinear if and only if the determinant: | a1 b1 c1 | | a2 b2 c2 | | a3 b3 c3 | equals zero. This means that the line through the given points a1:b1:c1 and a2:b2:c2 is found by points x:y:z that satisfy: | x y z | | a1 b1 c1 | = 0  | a2 b2 c2 | This can also be written as: (b1c2-b2c1) x + (c1a2-c2a1) y + (a1b2-a2b1) z = 0  From this we find the general appearance of a line formed by points x:y:z that satisfy: ax + by + cz = 0 for some a,b,c that are not all zero. There is a lot of similarity in computations with lines and with points: Three lines with coefficients a1,b1,c1 a2,b2,c2 and a3,b3,c3 are collinear if and only if the determinant: | a1 b1 c1 | | a2 b2 c2 | = 0 (Cf. ) | a3 b3 c3 | The point of intersection of two lines a1x+b1y+c1z = 0 and a2x+b2y+c3z = 0 is given by: b1c2-b2c1 : c1a2-c2a1 : a1b2-a2b1 (Cf. ) Barycentrics t:u:v are called "normalized" if t+u+v = 1. Normalized barycentrics have a lot of interesting properties, especially when one wants to calculate actual distances, areas, etc. Among the nice features of normalized barycentrics is also that one can use them to calculate triangle centers. For instance, the normalized barycentrics of the centroid are always given by 1/3:1/3:1/3. When we know the coordinates (barycentric, trilinear or even cartesian) of a triangle we can use the barycentrics to calculate the actual point. Let A(1,5), B(3,8), and C(5,-1) form the reference triangle. Then the point P with normalized barycentrics t:u:v is given by: P(t*1 + u*3 + v*5, t*5 + u*8 + v*(-1)) So the centroid of ABC is P(3,4). Barycentrics of triangle centers are usually given as functions of sidelengths a, b and c and/or angles A, B and C. As an example, the incenter of a triangle ABC always has barycentrics a:b:c. Triangle centers can be found - given in trilinears - at: http://faculty.evansville.edu/ck6/tcenters/index.html I hope this has helped. If you have a math question again, please send it to Dr. Math! Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/
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